`intcos^(3)xe^(log(sinx))dx` is equal to

Options

A. `-(sin^(4)x)/(4)+C`
B. `-(cos^(4)x)/(4)+C`
C. `(e^(sinx))/(4)+C`
D. none of these

Answer

To solve the integral \( \int \cos^3 x \, e^{\log(\sin x)} \, dx \), we can follow these steps:

### Step 1: Simplify the Integral
We start by simplifying the expression \( e^{\log(\sin x)} \). According to the properties of logarithms and exponents, we have:
\[ e^{\log(\sin x)} = \sin x \]
Thus, the integral becomes:
\[ \int \cos^3 x \, \sin x \, dx \]

### Step 2: Use Substitution
Next, we will use substitution to simplify the integral further. Let:
\[ t = \cos x \]
Then, the derivative of \( t \) with respect to \( x \) is:
\[ dt = -\sin x \, dx \quad \Rightarrow \quad \sin x \, dx = -dt \]
Now, we can rewrite the integral in terms of \( t \):
\[ \int \cos^3 x \, \sin x \, dx = \int t^3 (-dt) = -\int t^3 \, dt \]

### Step 3: Integrate
Now we can integrate \( -\int t^3 \, dt \):
\[ -\int t^3 \, dt = -\left( \frac{t^{4}}{4} \right) + C = -\frac{t^4}{4} + C \]

### Step 4: Substitute Back
Now, we substitute back \( t = \cos x \):
\[ -\frac{t^4}{4} + C = -\frac{\cos^4 x}{4} + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int \cos^3 x \, e^{\log(\sin x)} \, dx = -\frac{\cos^4 x}{4} + C \]